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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    A capacitor of capacity 10 \[\mu \]F is charged to a potential of 400 Volt. When its both plates are connected by a conducting wire, then heat generated will be

    A)  \[80\text{ }J\]             

    B)  \[0.8\text{ }J\]

    C)  \[8\times {{10}^{-3}}\]         

    D)   \[8\times {{10}^{-6}}J\]

    Correct Answer: A

    Solution :

     Given, the capacity of capacitor, \[C=10\times {{10}^{-6}}\mu F\] Potential\[V=400\]volt Stored energy = Heat generated \[U=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}(10\times {{10}^{-6}})\times {{(400)}^{2}}\] \[=\frac{1}{2}\times 10\times {{10}^{-6}}\times 16\times {{10}^{4}}=0.8J\]


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